Question

Which of the following have same radius according to Bohr's theory : (A) Radius of \(1^{st}\) orbit of H-atom.
(B) Radius of \(1^{st}\) orbit of He\(^{+}\).
(C) Radius of \(II^{nd}\) orbit of He\(^{2+}\).
(D) Radius of \(II^{nd}\) orbit of Li\(^{2+}\).
(E) Radius of \(II^{nd}\) orbit of Be\(^{3+}\).

• A. A and B
• B. A and E
• C. B, C and E
• D. A and D

Answer 1

The Correct Option is B

Concept: According to Bohr's model, the radius of the \(n^{th}\) orbit of a hydrogen-like species is given by \[ r_n = 0.529\,\text{\AA} \frac{n^2}{Z} = a_0 \frac{n^2}{Z} \] where \(a_0 = 0.529\,\text{\AA}\) (Bohr radius), \(n\) = principal quantum number, \(Z\) = atomic number. Two species will have the same radius if \( \frac{n^2}{Z} \) is the same.
Step 1: Calculate radius for each option. (A) \[ r = a_0\left(\frac{1^2}{1}\right) = a_0 \] (B) \[ r = a_0\left(\frac{1^2}{2}\right) = \frac{a_0}{2} \] (C) \[ r = a_0\left(\frac{2^2}{2}\right) = 2a_0 \] (D) \[ r = a_0\left(\frac{2^2}{3}\right) = \frac{4a_0}{3} \] (E) \[ r = a_0\left(\frac{2^2}{4}\right) = a_0 \]
Step 2: Compare the radii. (A) \(= a_0\) (E) \(= a_0\) Hence, both have the same radius. \[ \boxed{\text{Correct option: A and E}} \]